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If (1+a)^(n)=.^(n)C(0)+.^(n)C(1)a+.^(n)C...

If `(1+a)^(n)=.^(n)C_(0)+.^(n)C_(1)a+.^(n)C_(2)a^(2)+ . . +.^(n)C_(n)a^(n)`, then prove that
`.^(n)C_(1)+2.^(n)C_(2)+3.^(n)3C_(3)+ . . .+n.^(n)C_(n)=n.2^(n-1)`.

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The correct Answer is:
`.^(n)C_(1)2.^(n)C_(2)+3.^(n)C_(3)+ . . .+n^(n)C_(n)`
`=n+2.(n(n-1))/(1.2)+3(n(n-1)(n-2))/(1.2.3)+ . . .+n.1`
`=n[1+(n-1)/(1)+((n-1)(n-2))/(1.2)+ . . .+1]`
`=n[1+(n-1)/(1)+((n-1)(n-2))/(1.2)+ . . .+1]`
`=n[.^(n-1)C_(0)+.^(n-1)C_(1)+.^(n-1)C_(2)+ . . .+.^(n-1)C_(n-1)]`
=n {sum of the coefficient of `(1+a)^(n-1)`}
`=n.2^(n-1)`
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